Posts Tagged ‘torque’

Last time we discussed how torque is created as a flywheel spins. This torque is a factor of the flywheel’smoment of inertia, which is dependent on how far the masses of the flywheel’s parts are located from its center of rotation. Today we’ll present a formula to compute how much horsepower is required to accelerate a flywheel. And here it is,

Horsepower Required to Accelerate a Flywheel

where, T is the torque created on the flywheel’s shaft in units of inch-pounds. The term n is the flywheel shaft’s speed of rotation in revolutions per minute, RPM. Horsepower, HP, is engineering shorthand for a unit of power equal to 6600 inch-pounds per second, and the number 63,025 is a constant needed to convert torque,T, and the spinning shaft’s rotations per minute, RPM, into horsepower units.

Torque is present whether the flywheel’s spin accelerates or decelerates. During acceleration torque is created, which contributes to the production of kinetic energy that’s stored inside the flywheel. When a flywheel’s spin decelerates, its mass experiences the effects of negative acceleration, and stored kinetic energy is released.

As we learned awhile back, horsepower is a function of torque in any moving machinery, including engines and flywheels. An engine must produce horsepower to accelerate a flywheel connected to its shaft. By the same token, when the engine’s horsepower output diminishes or stops, the flywheel begins to decelerate. This deceleration causes kinetic energy stored within the flywheel to be released, providing horsepower necessary to keep the engine and flywheel spinning. That is, until the power output of the engine returns or the stored kinetic energy of the flywheel is ultimately exhausted.

We’ll see how that works next time when we take a look inside a reciprocating engine.

Last time we began our discussion on moment of inertia and how it affects a flywheel’s storage of kinetic energy. That inertia is a function of the flywheel’smass, in particular how the mass is distributed. Today we’ll continue our discussion and see how an engineering principal known as torque affects things.

Flywheel Torque and Distribution of Mass

We learned in a previous blog that torque is most simply defined as a measure of how much force acts upon an object to cause it to rotate around a pivot point or center of rotation, shown as a small black dot in the illustration. For our discussion we’ll focus on two parts of the flywheel, the hub, part A, and the rim, part E.

Part A has a massmA located a distance rA from the flywheel’s center, while part E has a mass mE located a distance rE from it. When an engine applies mechanical power to the flywheel by way of its rotating shaft, the revolutions per minute, RPM, increase and along with it the angular velocity, ω, also increases. For a refresher on this, follow the link.

As the flywheel’s angular velocity increases or decreases in response to the engine’s energy output, parts A and E reflect acceleration or deceleration of aA and aE. Since parts A and E exhibit both mass and acceleration, they are subject to Newton’s Second Law of Motion, which states that force equals mass times acceleration. Using that relationship we can calculate the force exerted on each part by,

FA = mA × aA (2)

FE = mE × aE (3)

Part A is small compared to part E, therefore mE is greater than mA and accordingly FE is greater than FA. Forces FA and FE act as torques, because they cause parts A and E to rotate around the flywheel’s center of rotation, so they are designated as Torque A, TA, and Torque E, TE. These torques are computed by,

TA = FA × rA (4)

TE = FE × rE (5)

Part E’s greater mass will contribute more torque than part A, and it will also contribute more to the flywheel’s kinetic energy content.

Most flywheels are designed with heavy rims supported by small hubs and slender spokes, because the more mass that’s distributed away from the flywheel’s center of rotation, the greater the flywheel’s moment of inertia and torque, and the more kinetic energy it can store.

Next time we’ll develop an equation which allows us to quantify the horsepower required to accelerate a flywheel.

We’ve been working towards a general understanding of how gear trains work, and today we’ll solve a final piece of the puzzle when we identify how increased gear train torque is gained at the expense of gear train speed.

Last time we developed a mathematical relationship between the torque, T, and the rotational speed, n, of the driving and driven gears in a simple gear train. This is represented by equation (8):

TDriven ÷ TDriving = nDriving÷ nDriven (8)

For the purpose of our example we’ll assume that the driving gear is mounted to an electric motor shaft spinning at 100 revolutions per minute (RPM) and which produces 50 inch pounds of torque.

Previous lab testing has determined that we require a torque of 100 inch pounds to properly run a piece of machinery that’s powered by the motor, and we’ve decided that the best way to get the required torque is not to employ a bigger, more powerful motor, but rather to install a gear train and manipulate its gear sizes until the desired torque is obtained. We know that using this approach will most likely affect the speed of our operation, and we want to determine how much speed will be compromised.

So if the torque on the driven gear needs to be 100 inch pounds, then what will be the corresponding speed of the driven gear?

To answer this question we’ll insert the numerical information we’ve been provided into equation (8). Doing so we arrive at the following:

TDriven÷ TDriving = nDriving÷ nDriven

(100 inch pounds) ÷ (50 inch pounds) = (100 RPM) ÷ nDriven

2 = (100 RPM) ÷ nDriven

nDriven= (100 RPM) ÷ 2 = 50 RPM

This tells us that in order to meet our torque requirement of 100 inch pounds, the gear train motor’s speed must be reduced from 100 RPM to 50 RPM, which represents a 50% reduction in speed, hence the tradeoff.

This wraps up our blog series on gears and gear trains. Next time we’ll move on to a new topic: Galileo’s experiments with falling objects.

In this blog series we’ve been examining gear train usefulness, specifically in terms of increasing torque. Equations presented last week began us on the final leg of our journey, and we’ve arrived at the point where the closing combination of equations will demonstrate the loss of speed that takes place when torque is increased within a gear train.

To that end, the two main equations under consideration as presented last week, are:

R = NDriven÷ NDriving = nDriving÷ nDriven

(1)

TDriving÷ TDriven = DDriving÷ DDriven

(2)

where R is the gear ratio of the gear train, N is the number of gear teeth, n is the gear rotational speed in revolutions per minute (RPM), T is the torque, and D is the gear pitch radius.

We were able to link these two equations by working through five key design equations applicable to simplified gear trains. For the full step-by-step progression see last week’s blog.

After working through the equations presented last time we were able to arrive at an equation which links equations (1) and (2). Here it is:

nDriven ÷nDriving= DDriving÷ DDriven

(7)

If you follow the color coding, you’ll see the elements of equations (1) and (2) which come together in equation (7). Because equation (7) links the gear speed ratios (red) with the gear pitch radii ratios (green), we can set the ratios in equation (1) equal to those in equation (2). Doing so, we get:

In order to see the tradeoff between speed and torque, we need only consider the parts of the equation which concern themselves with factors relating to speed and torque. Removing the other unnecessary factors, we arrive at:

TDriven ÷ TDriving = nDriving ÷ nDriven

(8)

Next week we’ll plug numbers into equation (8) and disclose the tradeoff of speed for torque.

Last time we used flashbacks to previous blogs in this series to revisit key equations in our ongoing discussion of gear trains and torque. We also introduced The Law of Conservation of Energy in conjunction with five equations that together demonstrate how when increasing torque by use of a simple gear train, we do so at the cost of speed.

Those five equations are:

R = NDriven÷ NDriving = nDriving÷ nDriven

(1)

TDriving÷ TDriven = DDriving÷ DDriven

(2)

TDriving = [HPDriving÷ nDriving] × 63,025

(3)

TDriven = [HPDriven÷ nDriven] × 63,025

(4)

HPDriving = HPDriven

(5)

where, R is the gear ratio, N the number of gear teeth, n the gear’s rotational speed, T the torque, D the gear pitch radius, and HP is the horsepower transmitted by the gears.

As we work the equations, keep in mind that our ultimate objective is to find a way to link together (1) and (2), the equations dealing with gear torque and speed. Once we accomplish this we’ll see how increased torque is obtained at the cost of speed. But because there are no common terms between equations (1) and (2), our first step is to develop one.

Developing a link between equations (1) and (2) is a process that begins with combining equations (2), (3), and (4).

TDriving÷ TDriven = DDriving÷ DDriven

(2)

TDriving = [HPDriving÷ nDriving] × 63,025

(3)

TDriven = [HPDriven÷ nDriven] × 63,025

(4)

The common terms in these three equations are TDriving and TDriven, so we’ll manipulate things in order to group them together. We’ll substitute equation (3) for the TDriving term in equation (2), and substitute equation (4) for the TDriven term in equation (2). We are now able to link all three equations to get:

{[HPDriving÷ nDriving] × 63,025} ÷ {[HPDriven÷ nDriven] × 63,025}

= DDriving÷ DDriven (6)

Now let’s go a step further to simplify equation (6). From equation (5) we know that the driving and driven gear horsepowers are equal. So, in equation (6), the HPDriving and HPDriven cancel out, along with the two 63,025 terms, allowing us to arrive at equation (7):

{[HPDriving÷ nDriving] ×63,025} ÷ {[HPDriven÷ nDriven] ×63,025}

= DDriving÷ DDriven

nDriven ÷nDriving = DDriving÷ DDriven

(7)

Next week we’ll use equation (7) to link together R, N, n, of equation (1) with D and T of equation (2) and in so doing disclose mathematically the tradeoff between torque and speed, then apply our findings to an example.

We’ve learned several methods to increase the torque of an electric motor through our series of articles on gear trains. One way is to attach a gear train to the motor’s shaft, a relatively simple thing to do. Today we’ll begin our exploration into how this method involves a tradeoff. It comes at the cost of speed.

We’ll begin our examination of the tradeoff at play by linking together key elements learned through past blogs on the subject of gear trains. We’ll revisit those lessons through flashbacks.

The first flashback we’ll make is to a blog entitled, Gear Ratio Formulas. There we learned that within a simple gear train consisting of two gears, the type most commonly employed to manipulate a motor’s torque, the ratio between the two gears, R, is relative to the ratio of their gear teeth, N. N is determined by the number of teeth each gear has in combination with the speeds, n, that each gear is going:

The tradeoff we’ve been alluding to comes in when gear speed, nDriven, represented in equation (1), is decreased, which results in an increase to TDriven in equation (2). But in order to see this we’ve got to somehow link the two equations together. In their present form there’s no common link between them. Or is there?

There actually is an indirect link between the two equations, which comes by way of the torque equation presented in another past blog. The third flashback we’ll make is to the blog discussing that subject, which is entitled, The Relationship Between Torque and Horsepower. Using facts presented in that blog, the torque equations for our two gears become:

TDriving = [HPDriving ÷ nDriving] × 63,025 (3)

TDriven = [HPDriven ÷ nDriven] × 63,025 (4)

Where’s the link between equations (1) and (2)? To answer that question we must reference a physics law known as The Law of Conservation of Energy . It states that the energy flowing from one gear to another within a gear train remains constant. Energy equates to horsepower, HP, in equations (3) and (4). So if the horsepower flowing through the gears is equal, our working equation becomes:

HPDriving = HPDriven (5)

Next time we’ll see how equation (5) is key to linking together equations (1) and (2) by way of equations (3) and (4). In so doing we’ll disclose the tradeoff to using gear trains.

Last time we learned that we can get more torque out of a motor by using one of two methods. In the first method we attach a gear train to the motor, then try different gear sizes until we arrive at the desired torque for the application. In the second method we eliminate the gear train altogether and simply use a higher horsepower motor to give us the torque we need.

Today we’ll explore the second method. We’ll use the equation presented in our last blog to determine torque, T, relative to a motor’s horsepower, HP, when the motor operatesat a speed, n:

T = [HP ÷ n] × 63,025

In earlier blogs in this series we employed a gear train attached to an electric motor to power a lathe. It provided an insufficient 200 inch pounds of torque when 275 is required. Let’s use the equation and a little algebra to see how much horsepower this motor develops if it turns at a speed of 1750 RPM, a common speed for alternating current (AC) motors:

200 inch pounds = [HP ÷ 1750 RPM] × 63,025

200 = HP × 36.01

HP = 200 ÷ 36.01 = 5.55 horsepower

For the purpose of our example today, let’s say we’ve nonsensically decided not to use a gear train, leaving us with no choice but to replace the underpowered motor with a more powerful one. So let’s see what size motor we’ll need to provide us with the required horsepower of 275 inch pounds.

Using the torque equation and plugging in numbers already provided our equation becomes:

275 inch pounds = [HP ÷1750 RPM] × 63,025

275 = HP × 36.01

HP = 275 ÷ 36.01 = 7.64 horsepower

This tells us that we need to replace the 5.55 horsepower motor with a 7.65 horsepower motor.

As you might have guessed, the higher the motor’s horsepower, the larger that motor’s size and weight — if you’re not using a gear train, that is. Bigger, bulkier motors cost more to purchase and operate and also take up more space, which often makes them impractical to use.

All this translates to the reality that sometimes it just makes more sense to use a gear train to provide more torque. It’s a lot easier and cheaper to attach a gear train to a motor and manipulate its gear sizes to arrive at desired torque than it is to buy a bigger motor.

You may have deduced by now that it’s relatively easy to get more torque. Almost too easy. Next time we’ll see how increased torque comes at another type of cost, the cost of speed.

We’ve been discussing gear trains for some time now, and last time we posed the question: Why even bother using a gear train and performing complex computations to arrive at a desired torque for an application? Why not just use a bigger motor to start with? Today we’ll see why.

First, we must acknowledge that sometimes higher torque is achieved by simply using a more powerful motor. But sometimes this isn’t possible or practical.

To begin our discussion, we must first understand how torque is related to motor power, the amount of mechanical work a motor can perform. Torque is in fact a function of how much mechanical power a motor produces. In the United States motor power is typically measured in units of horsepower.

The following equation illustrates the relationship between torque, horsepower, and motor speed:

T = [HP ÷ n] × 63,025

where T is the motor shaft’s torque in units of inch-pounds, HP is the motor’s horsepower, and n is the speed of the motor shaft in revolutions per minute (RPM). The number 63,025 in the equation is a constant used to convert the units of horsepower and RPM into units of torque (inch pounds). This equation applies to all sources of mechanical power. Its versatility enables design engineers to easily determine if a mechanical power source can deliver the torque required to drive a particular piece of machinery.

The torque equation above tells us that in order to get a higher torque T for a given speed n, you’ll have to get a motor with a higher HP. Put another way, if your speed remains constant and you use a motor with higher horsepower, you’ll get more torque for your application simply by increasing the horsepower.

Next time we’ll plug numbers into our equation and see how it all works.

Last week we worked with a gear train equation and found that the gears under consideration were not sized properly to run a lathe. Today we’ll increase the gear train torque and solve that problem.

How do we manipulate things to obtain the 275 inch pounds of torque required to drive the lathe? Last week we tried using a driven gear with a diameter of 8 inches and found that to be insufficient in size. So today the first thing we’ll try is a bigger driven gear, one with a pitch diameter of 8.5 inches. That’s 0.5 inches larger in diameter than the gear used in last week’s equation, and this just so happens to be the next size up in the gear manufacturer’s catalog.

As we did last week, we’ll begin our calculations with the torque ratio equation:

T1 ÷ T2 = D1 ÷ D2

We’ll use the same values as last week for T1, and D1, 200 inch pounds and 3 inches respectively, but we’ll increase the new value for D2, the driven gear pitch radius, to 4.25 inches (the new pitch diameter divided by two).

Inserting these values into the torque equation, the only variable remaining without a value is torque T2. Let’s determine that value now by using algebra to rearrange terms.

(200 inch pounds) ÷ T2 = (3 inches) ÷ (4.25 inches)

(200 inch pounds) ÷ T2 = 0.70

T2 = (200 inch pounds) ÷ (0.70) = 283.33 inch pounds

The value of T2 is found to be 283.33 inch pounds, which meets the torque requirement required to run the lathe. We were able to arrive at this torque by simply increasing the size of the driven gear relative to the size of the driving gear.

In the world of Newtonian physics, this is a rather straightforward arrangement. It all boils down to this simple dynamic: When the motor’s force is acting upon a wider gear, the force is located a longer distance from the center of the driving gear shaft, which results in more torque on the shaft.

As borne out by the example provided today, the larger the driven gear is in comparison to the driving gear, the more the gear train amplifies the torque that’s delivered by the motor. The principle at play here is exactly the same as that presented in a previous blog article where, for a given force exerted upon a wrench, torque was increased by simply increasing the length of the wrench handle.

Some of you may be wondering why we didn’t just use a bigger, more powerful motor to begin with, thereby eliminating the need for a gear train and all the calculations we’ve been running? We’ll see why that’s not always possible or practical next time.

Last time we learned that gear trains are torque converters, and we developed a torque ratio equation which mathematically ties the two gears in a gear train together. That equation is:

T1 ÷ T2 = D1 ÷ D2

Engineers typically use this equation knowing only the value for T2, the torque required to properly drive a piece of machinery. That knowledge is acquired through trial testing during the developmental phase of manufacturing.

Once T2 is known, a stock motor is selected from a catalog with a torque value T1 which closely approximates that of the required torque, T2. Then calculations are performed and lab tests are run to determine the driving and driven gear sizes, D1 and D2 which will enable the gear train to convert T1 into the required value of T2. This series of operations are often a time consuming and complex process.

To simplify things for the purpose of our example, we’ll say we’ve been provided with all values required for our equation, except one, the value of T2. In other words, we’ll be doing things in a somewhat reverse order, because our objective is simply to see how a gear train converts a known torque T1 into a higher torque T2.

We’ll begin by considering the gear train illustration above. For our purposes it’s situated between an electric motor and the lathe it’s powering. The motor exerts a torque of 200 inch pounds upon the driving gear shaft of the lathe, a torque value that’s typical for a mid sized motor of about 5 horsepower. As-is, this motor is unable to properly drive the lathe, which is being used to cut steel bars. We know this because lab testing has shown that the lathe requires at least 275 inch pounds of torque in order to operate properly.

Will the gears on our gear train be able to provide the required torque? We’ll find out next time when we insert values into our equation and run calculations.